Question

Calculus Disk Method problem - is there anyone who can check my work?

Answer 1

Suppose R is the region bounded by y = x^2, x = 2, and y = 0. A solid is generated by revolving R about the x-axis. Find the volume of the solid.

V = Pi * integral from 2 to 0 of [x^2]2 dx
V = Pi * integral from 2 to 0 of [x^4] dx
V = Pi * integral from 2 to 0 of [(x^5)/5]
V = Pi * [32/5] = 32Pi / 5 <--final answer

Answer 2

\(\large \color{red}{\checkmark}\)

Answer 3

No mistakes? I did get it right?

Answer 4

You're integrating FROM 0, to 2, yes?

Answer 5

Oh, yes.

Answer 6

yes u did get it right !

but why did u pick the interval from 2 to 0 , instead of 0 to 2 ?

Answer 7

it doesnt matter anyways.. .

Answer 8

just asking..

Answer 9

I think I wrote that backwards...sorry!

Answer 10

heh :D

Answer 11

This is what I meant. haha :)

Answer 12

ah you guys both wrote faster than me!

Answer 13

Haha thank you everyone! @ganeshie8 , @jeremyggg , @zepdrix !

Answer 14

haha okay, \(V = \Big|\int \limits_0^2 \pi y^2 dx\Big|\)

when calculating volumes of revolutions, we dont pay much attention to, whether we're doing 0->2 ro 2->0, unless ur prof is particular about it... abs value takes care of it

Answer 15

u wlc :)

Answer 16

Got it! :) @ganeshie8

Answer 17

@Kainui , please continue what you are writing if you'd like. I closed the question to let everyone know I've been helped. :)

Answer 18

But I'm still here to listen to what you have to say. :)

Answer 19

The idea behind it is simple: You're finding volume, which is simply: length*width*height right?

So if you can find a general formula for the volume of any thin slice in your shape, then you're fine.

So since Area is length times width, you have area of a circle as your cross section which is:

\[A=\pi*r^2\]

And they say "rotate" but really you're just using the function as the radius of any cross section right? So you just plug it in to get
\[A=\pi [f(x)]^2\]

So now that's the area of any cross section along your shape! Easy peasy. But remember that's area, we need volume. So we give it an infinitely thin volume we call dx

\[dV=A*dx=\pi [f(x)]^2 dx\]

So that's an infinitely small amount of volume. If only there was some... "sum" to add up an infinite number of infinitely small things between two points we could then find the total volume of all these slices. ;P

Answer 20

When I said, "So we give it an infinitely thin volume we call dx" I mean that dx is a length multiplied by an area to make an infinitely thin volume.. So A*dx is the same as length*width*height where as Area is just length times width. So that's how we got to volume. I hope that makes it a little more clear if it didn't already make sense.

Answer 21

@Kainui Wow, thanks for the explanation. We did learn about cross sections too, but your explanation was way better than what's in the textbook. :)

Answer 22

Yeah, I really think this is one of the most fun parts of calculus that's never explained... it's just like.. here's a formula... GO! lol

Answer 23

Haha, you are so right!! All they give out are formulas and some intricate explanation that doesn't even make sense (well, personally haha).

Answer 24

I wish i would have read that before i took my test. i am also in cal2 right now! we did that a while ago. now im trying to understand trig substitution... i guess you do not have an easy way to explain this huh kainui?

Answer 25

At least you did now :) and so did I. I think kainui left now.

Answer 26

Yeah I love trig sub and it's really not that bad once you understand it.

Answer 27

the particular part im lost on is \[\int\limits_{?}^{?} \frac{ dx }{ \sqrt{9+x ^{2}} }\]
im not quite sure what this has to do with trig substitution

Answer 28

There are really only a couple things to notice:

\[\sin^2\theta +\cos^2 \theta =1\]

simple old pythagorean identity. But you can manipulate it to get some other useful things. Why you need this is simple. Take the above identity and subtract sin^2theta from both sides to get:

\[\cos^2 \theta = 1-\sin^2 \theta\]

That sort of form is the key to simplifying stuff like this:

\[\sqrt{1-x^2}\]

Since the square root of that doesn't simplify nicely, we're stuck. But if we make x=sin(theta) then we have:

\[\sqrt{1-\sin^2 \theta}=\sqrt{\cos^2 \theta} = \cos \theta\]

The point is just to simplify square roots. The reason you might choose x=sec(theta) or x=tan(theta) is if the thing under the square root looks like:

x^2-1 or 1+x^2. So it's just a difference of negative sign placements.

Also, if you can't remember what the formula is for sec and tan just notice that if you take the pythagorean identity and divide both sides by cos^2(theta) you get it!

\[\frac{ \sin^2 \theta }{ \cos^2 \theta }+\frac{ \cos^2 \theta }{ \cos^2 \theta}=\frac{ 1}{ \cos^2 \theta}\]
same thing, no problem!
\[\tan^2 \theta + 1 = \sec^2 \theta\]

And of course if you are given something like:

\[\sqrt{x^2-9}\] then you just choose x = 3sec(theta) because that will give you\[\sqrt{9\sec^2 \theta - 9}=\sqrt{9(\sec^2 \theta -1)}=3\sqrt{\tan^2 \theta}=3\tan \theta\]

I hope that makes it clearer if you have questions I'm here. =P

Answer 29

I didn't notice you had just asked that question. Try it out and I'll help you figure it out if you need help.

Answer 30

thanks! i have been stuck on that one for about an hour and a half!

Answer 31

i think that i actually understand now :)

Answer 32

Awesome, glad to hear it. =)

Answer 33

i am not totally sure. the answer in the book is \[\ln \left| \sqrt{9+x ^{2}}+x \right|+c\] and i didnt finish but... that doesn't appear to be the way it is going to turn out

Answer 34

That looks about right to me. Don't forget that when you pick:

\[x=3\tan \theta\]
that means \[\frac{ dx }{ d \theta } = 3\sec^2 \theta\]

so multiply both sides by d theta to get:

\[dx = 3\sec ^2 \theta d \theta\] so plugging back into the integral we'll get an integral in terms of theta. But we have to convert back since we'll end up with a bunch of stuff. But just use your original substitution to figure it out! Solve here, and then you can just see that something you evaluate by looking at the triangle. It should work out, try it out.